UGC NET CSE | December 2013 | Part 3 | Question: 46

Question

A client-server system uses a satellite network, with the satellite at a height of 40,000 kms. What is the best-case delay in response to a request? (Note that the speed of light in air is 3,00,000 km/second)

• A. 133.33 m sec
• B. 266.67 m sec
• C. 400.00 m sec
• D. 533.33 m sec

Answer 1

The request has to go up and come down, and the response has to go up and come down.
The total path length traversed is thus 160,000 km.
The speed of light in air and vacuum is 300,000 km/sec,
So the propagation delay alone is 160,000/300,000 sec = 533.33 msec.

D is ans.

Comments

Hi,Can you please explain how you are getting traversed is thus 160,000 km?Is it not  40,000 kms *2=80,000kms?Please help me to understand

---

request from client to satellite

then  request from satellite to server

then response from server to satellite

then response from satellite to client

so total 40000*4

---

Thank you.yes answer should be D

---

@Prashant

Why a satellite will request client and client will send response to Satellite.. Client will only request to Server (satellite) and satellite will only send response. So the total ditance should be 80000km and final answer should be option B. Isnt?